解由x^2+2x+2=x^2+2x+1+1=(x+1)^2+1≥1即x^2+2x+2≥1两边去以1/3为底的对数则log1/3(x^2+2x+2)≤log1/3(1)=0即y≤0故函数的值域为（负无穷大，0].
